Lektion 1

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Elementtyper 3D
Eksempel 1 K 1 D 1 D 2
Page 1 and 2: FEM M5 Introduktion til FEM Søren
Page 4 and 5: Elementtyper 1D 1 frihedsgrad
Page 6 and 7: Elementtyper 2D
Page 10 and 11: Kx=F [k]{d}={R} K [Stivhedsmatrice]
Page 12 and 13: Løsning eksempel 1
Page 14 and 15: Matrice eksempel 2 R 1 R 2 R 3 k 1
Page 16: • Løsning eksempel 2
Page 19 and 20: Diskretisering af model Diskretiser
Page 21 and 22: Mesh 21
Page 23 and 24: Bjælkeelement. DOF = 23
Page 25 and 26: Stivhedsmatricen ∙ u 1 v 1 θ 1 u
Page 27 and 28: Stivhedsmatricen q 1 q 2 u 1 u 2 v
Page 29 and 30: Stivhedsmatricen http://people.civi
Page 35 and 36: Stivhedsmatricen 35
Page 41 and 42: Stivhedsmatricen q 1 q 2 N 1 N 2 V
Page 43 and 44: Stivhedsmatricen u 1 q 1 q 4 u q u
Page 57 and 58: Eksempel 1 2 3 P 0 D = 0 0 0 5
Page 59 and 60:
Eksempel 59 X X X X X X 0 0 0 X X X
Page 61 and 62:
Eksempel 61 X X X X 0 X X 0 0 X X X
Page 63 and 64:
FEM BM5 kursusgang Introduktion til
Page 65 and 66:
Modal analyse Beregningsmetoden giv
Page 67 and 68:
Fjeder/masse system u = u t K u = d
Page 69 and 70:
Fjeder/masse system u = u sin (ωt)
Page 71 and 72:
Fjeder/masse system u = u sin (ωt)
Page 73 and 74:
Fjeder/masse system f = ma f = mu K
Page 75 and 76:
Fjeder/masse system f = ma f = mu K
Page 77 and 78:
Fjeder/masse system c = 0 f = 0 K f
Page 79 and 80:
Fjeder/masse system 0 = mu + ku K m
Page 81 and 82:
Fjeder/masse system 0 = mu + ku K u
Page 83 and 84:
Fjeder/masse system 0 = −muω 2 s
Page 85 and 86:
Fjeder/masse system 0 = −muω 2 s
Page 87 and 88:
Dæmpning c = 0 C cr = 2m k m c < c
Page 89 and 90:
Dæmpning c = 0 ξ < 0,15
Page 91 and 92:
Dæmpning c = 0 ξ < 0,15 ω d = ω
Page 93 and 94:
Dæmpning c = 0 ξ < 0,15 ω d = 30
Page 95 and 96:
Matrise system D = D sin (ωt) D =
Page 97 and 98:
Page 99 and 100:
Page 101 and 102:
Mass Consistent mass metrix 0 L [N]
Page 103:
Eksempel ω 1 = 3,533 ω 2 = 34,81
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Lektion 1

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